Integrand size = 19, antiderivative size = 81 \[ \int \frac {1}{x \left (-a+b (c x)^n\right )^{5/2}} \, dx=-\frac {2}{3 a n \left (-a+b (c x)^n\right )^{3/2}}+\frac {2}{a^2 n \sqrt {-a+b (c x)^n}}+\frac {2 \arctan \left (\frac {\sqrt {-a+b (c x)^n}}{\sqrt {a}}\right )}{a^{5/2} n} \]
-2/3/a/n/(-a+b*(c*x)^n)^(3/2)+2*arctan((-a+b*(c*x)^n)^(1/2)/a^(1/2))/a^(5/ 2)/n+2/a^2/n/(-a+b*(c*x)^n)^(1/2)
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x \left (-a+b (c x)^n\right )^{5/2}} \, dx=\frac {2 \left (\frac {\sqrt {a} \left (-4 a+3 b (c x)^n\right )}{\left (-a+b (c x)^n\right )^{3/2}}+3 \arctan \left (\frac {\sqrt {-a+b (c x)^n}}{\sqrt {a}}\right )\right )}{3 a^{5/2} n} \]
(2*((Sqrt[a]*(-4*a + 3*b*(c*x)^n))/(-a + b*(c*x)^n)^(3/2) + 3*ArcTan[Sqrt[ -a + b*(c*x)^n]/Sqrt[a]]))/(3*a^(5/2)*n)
Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {891, 27, 798, 61, 61, 73, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x \left (b (c x)^n-a\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 891 |
\(\displaystyle \frac {\int \frac {1}{x \left (b (c x)^n-a\right )^{5/2}}d(c x)}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1}{c x \left (b (c x)^n-a\right )^{5/2}}d(c x)\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int \frac {1}{c x \left (b (c x)^n-a\right )^{5/2}}d(c x)^n}{n}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {-\frac {\int \frac {1}{c x \left (b (c x)^n-a\right )^{3/2}}d(c x)^n}{a}-\frac {2}{3 a \left (b (c x)^n-a\right )^{3/2}}}{n}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {-\frac {-\frac {\int \frac {1}{c x \sqrt {b (c x)^n-a}}d(c x)^n}{a}-\frac {2}{a \sqrt {b (c x)^n-a}}}{a}-\frac {2}{3 a \left (b (c x)^n-a\right )^{3/2}}}{n}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {-\frac {-\frac {2 \int \frac {1}{\frac {c^2 x^2}{b}+\frac {a}{b}}d\sqrt {b (c x)^n-a}}{a b}-\frac {2}{a \sqrt {b (c x)^n-a}}}{a}-\frac {2}{3 a \left (b (c x)^n-a\right )^{3/2}}}{n}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {-\frac {-\frac {2 \arctan \left (\frac {\sqrt {b (c x)^n-a}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {b (c x)^n-a}}}{a}-\frac {2}{3 a \left (b (c x)^n-a\right )^{3/2}}}{n}\) |
(-2/(3*a*(-a + b*(c*x)^n)^(3/2)) - (-2/(a*Sqrt[-a + b*(c*x)^n]) - (2*ArcTa n[Sqrt[-a + b*(c*x)^n]/Sqrt[a]])/a^(3/2))/a)/n
3.7.69.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_)*(x_))^(n_))^(p_.), x_Symbol] :> Simp[1/c Subst[Int[(d*(x/c))^m*(a + b*x^n)^p, x], x, c*x], x] /; FreeQ[{a , b, c, d, m, n, p}, x]
Time = 1.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {-\frac {2}{3 a \left (-a +b \left (c x \right )^{n}\right )^{\frac {3}{2}}}+\frac {2}{a^{2} \sqrt {-a +b \left (c x \right )^{n}}}+\frac {2 \arctan \left (\frac {\sqrt {-a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}}{n}\) | \(65\) |
default | \(\frac {-\frac {2}{3 a \left (-a +b \left (c x \right )^{n}\right )^{\frac {3}{2}}}+\frac {2}{a^{2} \sqrt {-a +b \left (c x \right )^{n}}}+\frac {2 \arctan \left (\frac {\sqrt {-a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}}{n}\) | \(65\) |
1/n*(-2/3/a/(-a+b*(c*x)^n)^(3/2)+2/a^2/(-a+b*(c*x)^n)^(1/2)+2/a^(5/2)*arct an((-a+b*(c*x)^n)^(1/2)/a^(1/2)))
Time = 0.29 (sec) , antiderivative size = 277, normalized size of antiderivative = 3.42 \[ \int \frac {1}{x \left (-a+b (c x)^n\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (2 \, \left (c x\right )^{n} \sqrt {-a} a b - \left (c x\right )^{2 \, n} \sqrt {-a} b^{2} - \sqrt {-a} a^{2}\right )} \log \left (\frac {\left (c x\right )^{n} b - 2 \, \sqrt {\left (c x\right )^{n} b - a} \sqrt {-a} - 2 \, a}{\left (c x\right )^{n}}\right ) + 2 \, {\left (3 \, \left (c x\right )^{n} a b - 4 \, a^{2}\right )} \sqrt {\left (c x\right )^{n} b - a}}{3 \, {\left (2 \, \left (c x\right )^{n} a^{4} b n - \left (c x\right )^{2 \, n} a^{3} b^{2} n - a^{5} n\right )}}, \frac {2 \, {\left (3 \, {\left (2 \, \left (c x\right )^{n} a^{\frac {3}{2}} b - \left (c x\right )^{2 \, n} \sqrt {a} b^{2} - a^{\frac {5}{2}}\right )} \arctan \left (\frac {\sqrt {\left (c x\right )^{n} b - a}}{\sqrt {a}}\right ) - {\left (3 \, \left (c x\right )^{n} a b - 4 \, a^{2}\right )} \sqrt {\left (c x\right )^{n} b - a}\right )}}{3 \, {\left (2 \, \left (c x\right )^{n} a^{4} b n - \left (c x\right )^{2 \, n} a^{3} b^{2} n - a^{5} n\right )}}\right ] \]
[-1/3*(3*(2*(c*x)^n*sqrt(-a)*a*b - (c*x)^(2*n)*sqrt(-a)*b^2 - sqrt(-a)*a^2 )*log(((c*x)^n*b - 2*sqrt((c*x)^n*b - a)*sqrt(-a) - 2*a)/(c*x)^n) + 2*(3*( c*x)^n*a*b - 4*a^2)*sqrt((c*x)^n*b - a))/(2*(c*x)^n*a^4*b*n - (c*x)^(2*n)* a^3*b^2*n - a^5*n), 2/3*(3*(2*(c*x)^n*a^(3/2)*b - (c*x)^(2*n)*sqrt(a)*b^2 - a^(5/2))*arctan(sqrt((c*x)^n*b - a)/sqrt(a)) - (3*(c*x)^n*a*b - 4*a^2)*s qrt((c*x)^n*b - a))/(2*(c*x)^n*a^4*b*n - (c*x)^(2*n)*a^3*b^2*n - a^5*n)]
Time = 4.69 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x \left (-a+b (c x)^n\right )^{5/2}} \, dx=\begin {cases} \frac {2 \left (- \frac {b}{3 a n \left (- a + b \left (c x\right )^{n}\right )^{\frac {3}{2}}} + \frac {b}{a^{2} n \sqrt {- a + b \left (c x\right )^{n}}} + \frac {b \operatorname {atan}{\left (\frac {\sqrt {- a + b \left (c x\right )^{n}}}{\sqrt {a}} \right )}}{a^{\frac {5}{2}} n}\right )}{b} & \text {for}\: b \neq 0 \\\frac {\log {\left (\left (c x\right )^{n} \right )}}{a^{2} n \sqrt {- a}} & \text {otherwise} \end {cases} \]
Piecewise((2*(-b/(3*a*n*(-a + b*(c*x)**n)**(3/2)) + b/(a**2*n*sqrt(-a + b* (c*x)**n)) + b*atan(sqrt(-a + b*(c*x)**n)/sqrt(a))/(a**(5/2)*n))/b, Ne(b, 0)), (log((c*x)**n)/(a**2*n*sqrt(-a)), True))
\[ \int \frac {1}{x \left (-a+b (c x)^n\right )^{5/2}} \, dx=\int { \frac {1}{{\left (\left (c x\right )^{n} b - a\right )}^{\frac {5}{2}} x} \,d x } \]
\[ \int \frac {1}{x \left (-a+b (c x)^n\right )^{5/2}} \, dx=\int { \frac {1}{{\left (\left (c x\right )^{n} b - a\right )}^{\frac {5}{2}} x} \,d x } \]
Timed out. \[ \int \frac {1}{x \left (-a+b (c x)^n\right )^{5/2}} \, dx=\int \frac {1}{x\,{\left (b\,{\left (c\,x\right )}^n-a\right )}^{5/2}} \,d x \]